The viscous drag acting on a metal sphere of diameter \(1\text{ mm}\), falling through a fluid of viscosity \(0.8\text{ Pa s}\) with a velocity of \(2\text{ m s}^{-1}\) is equal to
\(1.5 \times 10^{-3} \text{ N}\)
\(20 \times 10^{-3} \text{ N}\)
\(15 \times 10^{-3} \text{ N}\)
\(30 \times 10^{-3} \text{ N}\)
Solution:
According to Stokes' law, \(F = 6\pi \eta r v\). Here \(r = 0.5 \times 10^{-3}\text{ m}\), \(\eta = 0.8\text{ Pa s}\), and \(v = 2\text{ m s}^{-1}\). Calculating, we get \( F = 6 \times 3.14 \times 0.8 \times 0.5 \times 10^{-3} \times 2 \approx 15 \times 10^{-3}\text{ N}\).
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